If the angle sums are 2pi/k1, 2pi/k2, … for integers k1, k2, … greater than 1, then the curvature sum-rule implies 4pi =2pi (1-1/k1)+2pi (1-1/k2)+ … But the only solution is the case where there are just four terms, and k1=k2=k3=k4=2. If you allowed for general rational multiples of 2pi there would be more solutions (cube, regular icosahedron, …). -Veit
On Jul 16, 2015, at 11:35 PM, James Propp <jamespropp@gmail.com> wrote:
I want to think about rolling polyhedra, and the polyhedra I want to think about rolling are polyhedra in which the sum of the angles at each vertex is a submultiple of 360 degrees.
For instance, given any triangle T, we can create a tetrahedron whose four faces are all congruent to T, with angle-sum 180 degrees at each vertex (there's a name for such tetrahedra but I forget what it is).
Are there other polyhedra in which the angles at each vertex sum to a submultiple of 360 degrees (not necessarily the same one at each vertex)?
Jim Propp _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun